32 - 34 Maxima and minima problems of a rectangle inscribed in a triangle

Problem 32
Find the dimension of the largest rectangular building that can be placed on a right-triangular lot, facing one of the perpendicular sides.

Solution:

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Area:
$A = xy$

From the figure:
$\dfrac{y}{a - x} = \dfrac{b}{a}$

$y = \dfrac{b}{a}(a - x)$

$A = x \dfrac{b}{a}(a - x)$

$A = bx - \dfrac{b}{a}x^2$

$\dfrac{dA}{dx} = b - \dfrac{2b}{a}x = 0$

$\dfrac{2b}{a}x = b$

$x = \frac{1}{2}a$

$y = \dfrac{b}{a}(a - \frac{1}{2}a)$

$y = \frac{1}{2}b$

Dimensions: ½ a × ½ b answer

Problem 33
A lot has the form of a right triangle, with perpendicular sides 60 and 80 feet long. Find the length and width of the largest rectangular building that can be erected, facing the hypotenuse of the triangle.

Solution:

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Area:
$A = xy$

By similar triangle:
$\cot \alpha = \dfrac{r}{x} = \dfrac{60}{80}$

$r = \frac{3}{4}x$

$\cot \beta = \dfrac{s}{x} = \dfrac{80}{60}$

$s = \frac{4}{3}x$

$r + y + s = 100$

$\frac{3}{4}x + y + \frac{4}{3}x = 100$

$y = 100 - \frac{25}{12}x$

Thus,
$A = x(100 - \frac{25}{12}x)$

$A = 100x - \frac{25}{12}x^2$

$\dfrac{dA}{dx} = 100 - \frac{50}{12}x$

$x = 100(\frac{12}{50}$

$x = 24 \, \text{ feet }$

$y = 100 - \frac{25}{12}(24)$

$y = 50 \, \text{ feet }$

Dimensions: 50 ft × 24 ft answer

Problem 34
Solve Problem 34 above if the lengths of the perpendicular sides are a, b.